Question Number 243 by 123456 last updated on 25/Jan/15
$$\mathrm{evaluate} \\ $$$$\underset{−\infty} {\overset{+\infty} {\int}}\frac{\mathrm{sin}\:{x}}{{x}}{dx} \\ $$
Answered by prakash jain last updated on 17/Dec/14
![Let us consider ∫_0 ^∞ ∫_0 ^∞ e^(−st) sin t dt ds ∫_0 ^∞ ∫_0 ^∞ e^(−st) sin t dt ds=∫_0 ^∞ (1/(1+s^2 ))ds=[arctan s]_0 ^∞ =(π/2)...(1) ∫_0 ^∞ ∫_0 ^∞ e^(−st) sin t dt ds=∫_0 ^∞ ∫_0 ^∞ e^(−st) sin t ds dt =∫_0 ^∞ ((sin t)/t) dt....(2) From (1) and (2) ∫_0 ^∞ ((sin t)/t) dt =(π/2) ∫_(−∞) ^∞ ((sin t)/t) dt=2∫_0 ^∞ ((sin t)/t) dt=π](https://www.tinkutara.com/question/Q252.png)
$$\mathrm{Let}\:\mathrm{us}\:\mathrm{consider}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:{e}^{−{st}} \mathrm{sin}\:{t}\:{dt}\:{ds} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:{e}^{−{st}} \mathrm{sin}\:{t}\:{dt}\:{ds}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{1}}{\mathrm{1}+{s}^{\mathrm{2}} }{ds}=\left[\mathrm{arctan}\:{s}\right]_{\mathrm{0}} ^{\infty} =\frac{\pi}{\mathrm{2}}…\left(\mathrm{1}\right) \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:{e}^{−{st}} \mathrm{sin}\:{t}\:{dt}\:{ds}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:{e}^{−{st}} \mathrm{sin}\:{t}\:{ds}\:{dt} \\ $$$$=\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{sin}\:{t}}{{t}}\:{dt}….\left(\mathrm{2}\right) \\ $$$$\mathrm{From}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right) \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{sin}\:{t}}{{t}}\:{dt}\:=\frac{\pi}{\mathrm{2}} \\ $$$$\underset{−\infty} {\overset{\infty} {\int}}\:\frac{\mathrm{sin}\:{t}}{{t}}\:{dt}=\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{sin}\:{t}}{{t}}\:{dt}=\pi \\ $$